If the radii of the circles `(x-1)^2+(y-2)^2+(y-2)^2=1` and `(-7)^2+(y-10)^2=4` are increasing uniformly w.r.t. time as 0.3 units/s and 0.4 unit/s, respectively, then at what value of `t` will they touch each other?

Answer

`t-10,90`

Explanation

The given circles are
`(x-1)^(2)+(y-2)^(2)=1` (1)
and `(x-7)^(2)+(y-10)^(2)=4` (2)
Let `A-= (1,2), B-=(7,10),r_(1)=1,r_(2)=2`
`AB-=10,r_(2)+r_(2)=3`.
Now, `AB gt r_(1)+r_(2).` Hence, the two circles are non-intersecting.
The radii of the two circles at time t `1+0.3 t ` and `2+ 0.4 t`. For the two circles to touch each other.
`AB^(2)+[(r_(1)+0.3 t ) +-(r_(2)+0.4t)]^(2)`
or `100=[(1+0.3t)+-(2+0.4t)]^(2)`
or `100=(3+0.7t)^(2), [(0.1)t+1]^(2)`

or `3+0.7t= +-10,0.1 t = +- 10`
`t=10,t=90` `[ :' t gt 0]`
The two circles will touch each other externally in 10 s and internally in 90s.